Question

tìm x biết

8x^2-2x=0

Answer 1

\(8x^2-2x=0\Rightarrow4x\left(x-2\right)=0.\)

\(\Rightarrow\orbr{\begin{cases}x=0\\\left(x-2\right)=0\end{cases}\Rightarrow x\in\left\{0;2\right\}}\)

Answer 2

\(8x^2-2x=0\)

\(\Leftrightarrow x\left(8x-2\right)=0\)

\(\Leftrightarrow\orbr{\begin{cases}x=0\\8x-2=0\end{cases}}\Leftrightarrow\orbr{\begin{cases}x=0\\x=\frac{1}{4}\end{cases}}\)

Answer 3

\(8x^2-2x=0\)

\(\Leftrightarrow2x\left(4x-1\right)=0\)

\(\Leftrightarrow\orbr{\begin{cases}2x=0\\4x-1=0\end{cases}}\Leftrightarrow\orbr{\begin{cases}x=0\\x=\frac{1}{4}\end{cases}}\)

\(\Rightarrow\orbr{\begin{cases}x=0\\x=\frac{1}{4}\end{cases}}\)

Answer 4

\(8\times^2-2\times=0\)

\(\Rightarrow2\times\left(4\times-1\right)=0\)

\(\Rightarrow\orbr{\begin{cases}\times=0\\4\times-1=0\end{cases}\Rightarrow}\orbr{\begin{cases}\times=0\\\times=\frac{1}{4}\end{cases}}\)

Answer 5

8x² - 2x = 0

<=> 2x(4x - 1) = 0

<=> \(\orbr{\begin{cases}2x=0\\4x-1=0\end{cases}}\)

<=> \(\orbr{\begin{cases}x=0\\x=\frac{1}{4}\end{cases}}\)

Vậy ...

Answer 6

\(8x^2-2x=0\)

\(\Leftrightarrow2x\left(4x-1\right)=0\)

\(\Leftrightarrow\orbr{\begin{cases}2x=0\\4x-1=0\end{cases}\Leftrightarrow\orbr{\begin{cases}x=0\\4x=1\end{cases}\Leftrightarrow}\orbr{\begin{cases}x=0\\x=\frac{1}{4}\end{cases}}}\)

Vậy\(x=0;\frac{1}{4}\)

Answer 7

\(8x^2-2x=0\)

\(2x\left(4x-1\right)=0\)

\(\Rightarrow\orbr{\begin{cases}2x=0\\4x-1=0\end{cases}}\)                        \(\Rightarrow\text{ }\orbr{\begin{cases}x=0\\4x=0+1=1\end{cases}}\)           \(\Rightarrow\orbr{\begin{cases}x=0\\x=\frac{1}{4}\end{cases}}\)

                         \(\Rightarrow\text{ }x\in\left\{0\text{ ; }\frac{1}{4}\right\}\)

Answer 8

Bài làm

8x² - 2x = 0

=> 2x( 4x - 1 ) = 0

=> 2x = 0 => x = 0

hoặc ( 4x - 1 ) = 0 => x = 1/4

Vậy x = 0 hoặc x = 1/4

# Học tốt #